Question

A bullet of mass 6.0g is fired horizontally into a
2.6kg wooden block at rest on a horizontal surface. The coefficient
of kinetic friction between the block and surface is 0.38. The
bullet comes to rest in the block, which moves 1.8m.

1)What is the speed of the block immediately after the bullet comes
to rest within it?

2)At what speed is the bullet fired?

Answer #1

M = 2.6 kg

uk = 0.38

d = 1.8 m

let the intial speed of bullet = vi

(1)

Using momentum conservation,

Initial Momentum = Final Momentum

m*vi = (m+M) * vf

6.0 * 10^-3 * vi = (6.0 * 10^-3 + 2.6) * vf -----1

Using energy conservation for block,

Initial Kinetic Energy = Energy lost in friction

1/2 * (m+M) * vf^2 = uk* (m+M) * g * d

1/2 * vf^2 = 0.38 * 9.8 * 1.8

**vf = 3.66 m/s**

(2)

Substituing in eq 1,

6.0 * 10^-3 * vi = (6.0 * 10^-3 + 2.6) * 3.66

**vi = 1590 m/s
speed is the bullet fired, vi = 1590 m/s**

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